In this case, we have 6 places to fill from a pool of 22 candidates. There is no difference in the places: they are all commissioners, so if candidate A is selected, the order does not matter, but there is only one candidate, so there is no repetition.
Then, this is a combination (as the order does not matter) with no repetition of 22 elements in 6 places.
We can calculate this as:
[tex]\begin{gathered} C(n,r)=\frac{n!}{r!(n-r)!} \\ C(22,6)=\frac{22!}{6!(22-6)!}=\frac{22!}{6!\cdot14!}=\frac{22\cdot21\cdot20\cdot19\cdot18\cdot17\cdot16\cdot15}{6\cdot5\cdot4\cdot3\cdot2\cdot1}=17907120 \end{gathered}[/tex]Answer: there are 17,907,120 ways this can be done.